The x and y coordinates of the particle at any time are `x=5t-2t^(2) and y=10t` respectively, where x and y are in meters and t in seconds. The acceleration of the particle at `t=2s` is:

Options

A. 0
B. `5m//s^(2)`
C. `-4m//s^(2)`
D. `-8m//s^(2)`

Answer

c

Explanation

Given, `x=5t-2t^(2)`
Velocity of the particle ,
`v_(x)=(dx)/(dt)=(d)/(dt)(5t-2t^(2))=5-4t`
Acceleration, `a_(x)=(d)/(dt)v_(x)=-4ms^(-2)`
Also, y = 10 t
Velocity, `v_(y)=(dy)/(dt)=10`
`:.` Net acceleration of the particle ,
`a_(net)=a_(x)hati+a_(y)hatj=(-4ms^(-2))hati`
or `a_(net) =4hatims^(-2)`